Problem: $2x^3-5xy-y^2=3$ Find $\dfrac{dy}{dx}$. Choose 1 answer: Choose 1 answer: (Choice A) A $\dfrac{3+2y}{6x^2-5}$ (Choice B) B $\dfrac{6x-2y}{5}$ (Choice C) C $\dfrac{6x^2-5y}{5x+2y}$ (Choice D) D $\dfrac{6x^2-5y}{2y-5x}$
We cannot isolate $y$ in order to define it as a function of $x$. Therefore, $2x^3-5xy-y^2=3$ defines $y$ as a function of $x$ implicitly. To find $\dfrac{dy}{dx}$, we need to perform implicit differentiation. In implicit differentiation, we differentiate both sides of the equation according to $x$, and treat $y$ as an implicit function of $x$. [I need more explanation about implicit differentiation!] $\begin{aligned} 2x^3-5xy-y^2&=3 \\\\ \dfrac{d}{dx}(2x^3-5xy-y^2)&=\dfrac{d}{dx}(3) \\\\ \dfrac{d}{dx}(2x^3)-5\dfrac{d}{dx}(xy)-\dfrac{d}{dx}(y^2)&=0 \\\\ 6x^2-5\Bigl(1\cdot y+x\cdot\dfrac{dy}{dx}\Bigr)-2y\cdot\dfrac{dy}{dx}&=0 \\\\ 6x^2-5y-5x\cdot\dfrac{dy}{dx}-2y\cdot\dfrac{dy}{dx}&=0 \end{aligned}$ Once we've completed the differentiation, we can arrange the equation so $\dfrac{dy}{dx}$ is isolated: $\begin{aligned} 6x^2-5y-5x\cdot\dfrac{dy}{dx}-2y\cdot\dfrac{dy}{dx}&=0 \\\\ \dfrac{dy}{dx}(-5x-2y)&=-6x^2+5y \\\\ \dfrac{dy}{dx}&=\dfrac{-(6x^2-5y)}{-(5x+2y)} \\\\ &=\dfrac{6x^2-5y}{5x+2y} \end{aligned}$ In conclusion, $\dfrac{dy}{dx}=\dfrac{6x^2-5y}{5x+2y}$.